Question

помлгитея!!!!!!!!!!!!!++!++++++

Answer 1

sin^3(2x)=(3/4)sin2x-(1/4)sin6x

$(3/4)*intlimits^a_b {sin2x} , dx -(1/4)*intlimits^a_b {sin6x} , dx$=

=-(3/4)cosx+(1/12)cos3x

ответ  (1/12)cos3x-(3/4)cosx